GATE | GATE-CS-2017 (Set 2) | Question 54
Consider a machine with byte addressable memory of 232 bytes divided into blocks of size 32 bytes. Assume a direct mapped cache having 512 cache lines is used with this machine. The size of tag field in bits is _____
Explanation: Total address space = 32 bit
Block size (B) = 5 bit
No. of bit to represent line no (L) = 9.
Tag bit + B + L = 32
==> X + 5 + 9 = 32
==> X = 18
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